Differential of exponential operator

[SK1.618]

If $\hat{U}(r) = e^{\hat{A}(r)}$, can we say $\frac{d\hat{U}}{dr} = \frac{d\hat{A}}{dr}e^{\hat{A}(r)}$?

 

[JorisL]

This expression can be used when $$\left[\frac{d \hat{A}}{dr},A\right] =0$$.

I think you should use the general Leibniz rule for products.
While checking this I found http://www.aps.anl.gov/Science/Publications/lsnotes/content/files/APS_1418211.pdf
p9-10 pertains to this question.

 

[micromass]

If $\hat{U}(r) = e^{\hat{A}(r)}$, can we say $\frac{d\hat{U}}{dr} = \frac{d\hat{A}}{dr}e^{\hat{A}(r)}$?

For matrices, this is true provided that $A$ is differentiable.

The general (infinite-dimensional) case is a bit more annoying, especially if you let $A(r)$ be unbounded. So since I think you're only really interested in the case $A(r) = irA$ with $A$ self-adjoint (which is actually a fairly general case: http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups ), then you can indeed say that

$$\frac{d U}{dr} = iA$$

in the sense that for each $\psi$ in the domain of $A$ holds that

$$\lim_{r\rightarrow 0} \left\|\frac{U(r)\psi - \psi}{r} - iA\psi\right\| = 0$$

Books like Reed & Simon, and Hall's QM for mathematicians give proofs. A naive proof of expanding the exponential in a power series would probably work in the case that $A$ is a bounded operator, but likely fails in the unbounded case. You need the spectral theorem there (you already need it just to define the exponential).